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\(M=1+\frac{1}{3}-\frac{1}{3^2}+\frac{1}{3^3}-\frac{1}{3^4}+......+\frac{1}{3^{19}}-\frac{1}{3^{20}}\)
\(\Rightarrow\frac{1}{3}M=\frac{1}{3}+\frac{1}{3^2}-\frac{1}{3^3}+\frac{1}{3^4}-\frac{1}{3^5}+.......+\frac{1}{3^{20}}-\frac{1}{3^{21}}\)
\(\Rightarrow\frac{1}{3}M+M=1+\frac{1}{3}+\frac{1}{3}-\frac{1}{3^{21}}\)
\(\Rightarrow\frac{4}{3}M=\frac{5}{3}-\frac{1}{3^{21}}\)\(\Rightarrow M=\frac{\frac{5}{3}-\frac{1}{3^{31}}}{\frac{4}{3}}\)
2A=1-1/2+1/2^2-...+1/2^98-1/2^99
=>3A=1-1/2^100
=>\(A=\dfrac{2^{100}-1}{3\cdot2^{100}}\)
a, ( 1/2 + 1) . ( 1/3 + 1) . (1/4 + 1) ... ( 1/999 + 1)
= 3/2 . 4/3 . 5/4 . 1000/999
= 1/2 . 1/1 . 1/1 ... 1000/1
= 1000/2
= 500
b, (1/2-1) . (1/3-1) . (1/4-1) ... (1/1000-1)
= -1/2 . (-2)/3 . (-3)/4 ... (-999)/1000
= (-1)/1 . (-1)/1 . (-1)/1 ... (-1)/1000
= (-1)/1000
c, 3/2^2 . 8/3^2 . 15/4^2 ... 99/10^2
= 1.3/2.2 * 2.4/3.3 * 3.5/4.4***9.11/10.10
=( 1.2.3...99).(3.4.5...11)/(2.3.4....10).(2.3.4...10)
= 1.11/2.10
= 11/20
Ta có :
\(B=\left(1-\frac{1}{2}\right)\left(1-\frac{1}{3}\right).....\left(1-\frac{1}{2003}\right)\left(1-\frac{1}{2004}\right)\)
\(B=\frac{1}{2}.\frac{2}{3}.....\frac{2002}{2003}.\frac{2003}{2004}\)
\(B=\frac{1.2.....2002.2003}{2.3.....2003.2004}\)
\(B=\frac{1}{2004}\)
Vậy \(B=\frac{1}{2004}\)
Chúc bạn học tốt ~
B = (1 + 1/2)(1 + 1/3)(1 + 1/4) ...(1 + 1/100)
= \(\frac{3}{2}.\frac{4}{3}.\frac{5}{4}...\frac{101}{100}\)
= \(\frac{3.4.5....101}{2.3.4...100}=\frac{101}{2}\)
C = \(\left(1-\frac{1}{2}\right)\left(1-\frac{1}{3}\right)\left(1-\frac{1}{4}\right)....\left(1-\frac{1}{1000}\right)\)
\(=\frac{1}{2}.\frac{2}{3}.\frac{3}{4}....\frac{999}{1000}\)
\(=\frac{1.2.3...999}{2.3.4....1000}=\frac{1}{1000}\)